| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Area using perpendicular distance |
| Difficulty | Standard +0.8 This is a multi-part coordinate geometry question requiring: (i) finding a line equation through two points (routine), (ii) calculating perpendicular distance between parallel lines using the formula (moderately challenging for C1), and (iii) applying parallelogram area = base × height. Part (ii) requires recognizing the lines are parallel, using the distance formula with specific form, and algebraic manipulation to reach the required surd form—this elevates it above standard C1 fare. The 13 total marks and proof element ('show that') indicate above-average difficulty. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=0 |
| Answer | Marks |
|---|---|
| (i) \(\text{grad} = \frac{4-3}{3-(-1)} = \frac{1}{4}\) | M1, A1 |
| Answer | Marks |
|---|---|
| \(x - 4y + 13 = 0\) | M1, A1 |
| (ii) \(\text{perp grad} = \frac{-1}{\frac{1}{4}} = -4\) | M1 |
| Answer | Marks |
|---|---|
| \(y = -4x - 1\) | M1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = 1, \therefore (1, -5)\) | M1, A1 | |
| dist. A to \((1, -5) = \sqrt{(1+1)^2 + (-5-3)^2} = \sqrt{4 + 64} = \sqrt{68}\) | M1 | |
| \(\therefore \text{dist. between lines} = \sqrt{68} = \sqrt{4 \times 17} = 2\sqrt{17}\) | A1 | \([k = 2]\) |
| (iii) \(AB = \sqrt{(3+1)^2 + (4-3)^2} = \sqrt{16 + 1} = \sqrt{17}\) | M1 | |
| area \(= \sqrt{17} \times 2\sqrt{17} = 34\) | A1 | (13 marks) |
**(i)** $\text{grad} = \frac{4-3}{3-(-1)} = \frac{1}{4}$ | M1, A1 |
$\therefore y - 3 = \frac{1}{4}(x + 1)$
$4y - 12 = x + 1$
$x - 4y + 13 = 0$ | M1, A1 |
**(ii)** $\text{perp grad} = \frac{-1}{\frac{1}{4}} = -4$ | M1 |
line through A, perp $l_1$: $y - 3 = -4(x + 1)$
$y = -4x - 1$ | M1, A1 |
intersection with $l_2$: $x - 4(-4x - 1) - 21 = 0$
$x = 1, \therefore (1, -5)$ | M1, A1 |
dist. A to $(1, -5) = \sqrt{(1+1)^2 + (-5-3)^2} = \sqrt{4 + 64} = \sqrt{68}$ | M1 |
$\therefore \text{dist. between lines} = \sqrt{68} = \sqrt{4 \times 17} = 2\sqrt{17}$ | A1 | $[k = 2]$
**(iii)** $AB = \sqrt{(3+1)^2 + (4-3)^2} = \sqrt{16 + 1} = \sqrt{17}$ | M1 |
area $= \sqrt{17} \times 2\sqrt{17} = 34$ | A1 | (13 marks)
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**Total: (72 marks)**\includegraphics{figure_9}
The diagram shows the parallelogram $ABCD$.
The points $A$ and $B$ have coordinates $(-1, 3)$ and $(3, 4)$ respectively and lie on the straight line $l_1$.
\begin{enumerate}[label=(\roman*)]
\item Find an equation for $l_1$, giving your answer in the form $ax + by + c = 0$, where $a$, $b$ and $c$ are integers. [4]
\end{enumerate}
The points $C$ and $D$ lie on the straight line $l_2$ which has the equation $x - 4y - 21 = 0$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Show that the distance between $l_1$ and $l_2$ is $k\sqrt{17}$, where $k$ is an integer to be found. [7]
\item Find the area of parallelogram $ABCD$. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR C1 Q9 [13]}}